假设表格中有10行。其中只有5个在屏幕上可见,表示当表加载时,1,2,3,4,5行可见。当我向下滚动1行然后第2行,3,4,5,6可见。在Javascript或jquery中是否有任何方法可以找到屏幕上的第一行?我尝试过可能的方式:
if(tr.position().top > $(table).position().top)
{
}
else
{
}
这里我比较每行的位置和桌面位置。但不幸的是,即使我滚动,每次我都获得相同的位置。怎么做?
答案 0 :(得分:1)
使用here中的代码进行一些修改:
JS代码(使用div
代替tr
):
$(function () {
$(window).scroll(function () {
var first = null; // element at the top
$("div").each(function(){ // check each div
// if it is visivle, and is there is not first element yet,
if( isScrolledIntoView($(this)) && !first) {
first = $(this); // this is the first
first.addClass("seen"); // visual aspect
}
// unnecessary, only for visual
else
$(this).removeClass("seen"); // remove class seen (visual)
});
});
// foreign code
function isScrolledIntoView(elem) {
// .....
}
});
<强> DEMO 强>
滚动查看效果
答案 1 :(得分:0)
你可以在滚动事件回调上做这样的事情(使用JQuery),使用超时来限制事件的频率,加上最后找到的可见元素的光标,以便更快地搜索下一个事件。这里&#39;这个&#39;是你观点的背景:
onScroll: function(e) {
if (this.scrollEvent) {
return;
}
this.scrollEvent = true;
setTimeout(function(view) {
// @replace you should check here is you view still exists using a status on your view context
if (view.isDestroyed) {
return;
}
view.scrollEvent = false;
var top = $('youselector on tbody').parent().parent().offset().top;
if (view.previousTopElement) {
if (view.previousTopElement.offset().top <= top) {
// the element is the previous, or on bottom of previous
var element = view.previousTopElement;
var i = view.previousTopElementIndex;
view.previousTopElement = null;
while (element.length) {
if (element.offset().top + element.height() >= top) {
// you can uses i + 1 for a 1 based position to display
view.previousTopElement = element;
view.previousTopElementIndex = i;
return;
}
element = element.next();
++i;
}
} else if (view.previousTopElement.offset().top > top) {
// the element is before the previous
var element = view.previousTopElement;
var i = view.previousTopElementIndex;
this.previousTopElement = null;
while (element.length) {
if (element.offset().top <= top) {
// you can uses i + 1 for a 1 based position to display
view.previousTopElement = element;
view.previousTopElementIndex = i;
return;
}
element = element.prev();
--i;
}
}
}
// no previous, get all rows
var rows = $('youselector on tbody').children('tr');
view.previousTopElement = null;
view.previousTopElementIndex = 0;
$.each(rows, function (i, el) {
var element = $(el);
if (element.offset().top + element.height() >= top) {
// you can uses i + 1 for a 1 based position to display
view.previousTopElement = element;
view.previousTopElementIndex = i;
return false;
}
});
}, 100, this); // you can adjust the timeout, pass your view context
}
这是个主意。在我的情况下,我做了这个优化,因为我可以有很多行。